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Relativistic Invariance

by Ed Quanta
Tags: invariance, relativistic
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turin
#109
Dec11-03, 08:01 PM
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Originally posted by selfAdjoint
orthonormal means "mutually perpendicular and of magnitude one". Like a set of unit vectors spanning the space. [tex] v_i v_j = \delta_j^i[/tex]
I'm assuming this is supposed to be a dot product? Don't you need to raise one of the indices to do that? I don't mean to be picky, but I was under the impression that you must contract covariant components with contravariant components, but that lethe found the notion of "perpendicular" distasteful. I just wanted to hear how "orthonormal" does not require a notion of "perpendicular."
lethe
#110
Dec11-03, 08:37 PM
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Originally posted by turin
I'm assuming this is supposed to be a dot product? Don't you need to raise one of the indices to do that?
yes, if he wants to use the einstein summation notation, one of those indices should be raised.

I don't mean to be picky, but I was under the impression that you must contract covariant components with contravariant components, but that lethe found the notion of "perpendicular" distasteful. I just wanted to hear how "orthonormal" does not require a notion of "perpendicular."
when you have a metric present, the notion of perpendicular is perfectly well acceptable. necessary even. two vectors are perpendicular if their dot product is zero.

but note that you have dual vectors independently of the metric, and that the metric does not allow you to take inner products with a vector and a dual vector, but rather only two vectors (or also two dual vectors).

so it makes no sense to talk about a vector and a dual vector being perpendicular, and that is what i was objecting to before.
Ambitwistor
#111
Dec11-03, 09:15 PM
P: 837
Originally posted by turin
They are not hard for me to visualize, even on a "chart" (I think that's the term I'm looking for, a piece of the manifold that is presented as a flat map). The issue I have with lat. & long. is that, if I look at a map of Scandanavia, for instance, I see this lat. & lon. coordinate system dramatically curved, but I see no good reason for it in the context of the chart.
Whether they look curved on a chart depends on the chart. If you choose a something like a Mercator projection, where latitude and longitude lines get mapped to straight lines in the Euclidean plane, then they'll look straight. If you don't, they won't.

But what does whether they look straight or curved have to do with your ability to visualize them?

I thought that a coordinate system had a set of basis vectors. I must be very confused about what a coordinate system is.
A coordinate system is a collection of functions on the manifold, each of which assigns a number (a coordinate) to a given point.

You don't need a coordinate system in order to define a tangent space, and you don't need a tangent space in order to define a coordinate system.

It specifies to me that Houston is 300 miles directly in front of me.
I was taking your statement rather literally: most road signs do not indicate that something is directly ahead of you, they just indicate how far away it is (and maybe a rough sense of direction).

How much more specific can one be about location? Specifying the distance in degrees of lattitude I find less specific, in the sense that I have a standard for the mile that will tell me how many times my tires will have to rotate before I get there, whereas I don't have a standard for a degree of lattitude that will tell me anything at all about the duration of my trip in terms of tire rotation or anything else.
Coordinates themselves don't tell you anything about distances. The metric tells you about distances.

So do I jump from one to another? Where am I in between tangent planes?
You're never "between tangent planes". You are always in some tangent plane, because you are always at some point, and every point has a tangent plane associated with it.
Ambitwistor
#112
Dec11-03, 09:19 PM
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Originally posted by lethe
why can t you just compute the integral curves of the basis vectors of the vielbein and use those to parametrize the space?
Does your question have anything to do with anything I said?
lethe
#113
Dec11-03, 09:38 PM
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Originally posted by Ambitwistor
Does your question have anything to do with anything I said?
i think so... you said

Originally posted by Ambitwistor
but it looks to me like you demonstrated that you can find an orthonormal basis at every point in the manifold (a frame field).
ok, we have a bunch of vectors in the tangent space at each point that are orthonormal. this is a vielbein, or frame field.
That's true, but you can't find a coordinate system on a curved space in which the metric components are the Kronecker delta everywhere.
right. a nonflat space shouldn t have a flat metric. i agree, but then i thought of a question...

my question is this: if i have this orthonormal basis at each point, can i not integrate them to find the flow? can i not parametrize along these flows? will this parametrization not give me back the Kronecker delta as the components of the metric in these coordinates?

that would be a contradiction if i were able to do that. so clearly it must not be possible. but on the other hand, i am under the impression that one can always integrate a vector field to find its flow.

so this is my question. i think it has something to do with what you said, but i guess if you don t think so, well.... i must be awfully confused about a lot of things.
Ambitwistor
#114
Dec11-03, 09:52 PM
P: 837
Originally posted by lethe
my question is this: if i have this orthonormal basis at each point, can i not integrate them to find the flow? can i not parametrize along these flows? will this parametrization not give me back the Kronecker delta as the components of the metric in these coordinates?
Take polar coordinates. The integral curves of the [itex]\vec{r}[/itex] and [itex]\vec{\theta}[/itex] vector fields are lines and circles, respectively. Now replace those fields with their normalized versions [itex]\hat{r}[/itex] and [itex]\hat{\theta}[/itex]. The [itex]\hat{r}[/itex] vector at a point is still tangent to the same line, and the [itex]\hat{\theta}[/itex] vector is still tangent to the same circle, so you could say that those curves are integral curves of the frame field. But if you try to build coordinates out of those curves by constructing their tangent vectors (derivative operators), you don't get the frame field back, you get back [itex]\vec{r}[/itex] and [itex]\vec{\theta}[/itex], which are coordinate vector fields. The frame field isn't a coordinate vector field; the vectors don't commute with each other. And thus when you try to find the components of the metric in that coordinate system, you're not going to get the Kronecker delta.
lethe
#115
Dec12-03, 08:34 AM
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Originally posted by Ambitwistor
Take polar coordinates. The integral curves of the [itex]\vec{r}[/itex] and [itex]\vec{\theta}[/itex] vector fields are lines and circles, respectively. Now replace those fields with their normalized versions [itex]\hat{r}[/itex] and [itex]\hat{\theta}[/itex]. The [itex]\hat{r}[/itex] vector at a point is still tangent to the same line, and the [itex]\hat{\theta}[/itex] vector is still tangent to the same circle, so you could say that those curves are integral curves of the frame field.
would you? i suspect two circles would be parametrized differently, and therefore have different magnitudes of the tangent vectors.

But if you try to build coordinates out of those curves by constructing their tangent vectors (derivative operators), you don't get the frame field back, you get back [itex]\vec{r}[/itex] and [itex]\vec{\theta}[/itex], which are coordinate vector fields.
by the very definition of integral curve of a vector, i must get that vector back as the tangent vector of the integral curve. this is the definition of integral curve, so if this isn t satisfied, this ain t an integral curve.

should i infer that these integral curves don t exist? can you explain to me how come this integral curve does not have the vector it is an integral curve of as tangent?

The frame field isn't a coordinate vector field; the vectors don't commute with each other.
sure, i understand this, but i don t see how this implies your other statements. are you going to say something like the tangent vectors of integral curves of noncommuting vectors fields are commuting vector fields? does noncommutativity somehow make the tangent of a flow of a vector be a different vector?

And thus when you try to find the components of the metric in that coordinate system, you're not going to get the Kronecker delta.
i am going to try a calculation, perhaps.
turin
#116
Dec12-03, 10:38 AM
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Originally posted by lethe
... dual vectors independently of the metric, and that the metric does not allow you to take inner products with a vector and a dual vector, but rather only two vectors (or also two dual vectors).
OK, now we've uncovered more of my confusion/inconsistency in the way I'm trying to view this stuff.

From QM, I remember that the inner product was a bra multiplying a ket (at least, that's how it was done in Shankar). I got the impression in QM that the bra and ket are literally different objects, even if it is the bra version and ket version of the same state. There was never a metric idea introduced, but orthogonality was still an important issue.

From classical mechanics, I remember that any vector has two different kinds of components and two different kinds of bases, contravariant and covariant. The contravariant basis is like arrows, and the covariant basis is like a set of planes (stack of pancakes). Taking a dot product (which I'm still trying to decide if this is really the same as an "inner product") of two vectors is summing the products of the contravariant components of one of them with the covariant components of the other. If the dot product is zero, I interpretted it as the arrow being parallel to the planes in the set of planes. The dot product is maximum when the arrow is perpendicular to the set of planes (so that it passes through the maximum number of planes).

Are you saying that my interpretations from classical mechanics are inappropriate? Should I think about it more in terms of the way I think about the inner product in QM?
turin
#117
Dec12-03, 10:55 AM
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Originally posted by Ambitwistor
A coordinate system is a collection of functions on the manifold, each of which assigns a number (a coordinate) to a given point.

You don't need a coordinate system in order to define a tangent space, and you don't need a tangent space in order to define a coordinate system.
After letting this soak overnight, I think I realize what y'all are saying. I don't know if I quite follow your set of functions description though. But now it does make sense to me that the points in the manifold exist, period. The coordinate system is a way to keep track of them, like assigning an index to the elements of a set? Since the points are not discrete elements, we can't use an index (at least not like i = 1,2,3,...), so we use a continuous variable and call it a coordinate. Just like I could index the elements of a set differently, I can use a different coordinate.

But now I am confused about why one needs two coordinates instead of one. Is there no way to use just one coordinate to indicate points. I am having trouble shifting my mind from elements of a set needing only one index and points in a manifold needing more than one coordinate. (I am at this point going to assume that my analogy was correct that points:manifold::elements:set) In a vector space, I've got the idea of linear independence dictating how many independent vectors I will need in my basis to span the vector space, so the dimension is clear. I don't see the same indicator in a manifold.

Can you explain the collection of functions a bit?




Originally posted by Ambitwistor
You're never "between tangent planes". You are always in some tangent plane, because you are always at some point, and every point has a tangent plane associated with it.
But I thought that the tangent planes for any two points, no matter how close they are, have absolutely nothing to do with each other. Unless I'm allowed to, in some sense, take the tangent plane with me, then I have to go from one to another. But, if adjacent tangent planes have nothing to do with each other, then this process seems disjoint, abrupt, or something. There just seems to be a discontinuity here, but I can't put my finger on it. I'll have to let this idea soak in my mind, too.
turin
#118
Dec12-03, 11:08 AM
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Originally posted by Ambitwistor
The integral curves of the [itex]\vec{r}[/itex] and [itex]\vec{\theta}[/itex] vector fields are lines and circles, ...
What vector fields? I only know of vector fields like the electric field and such. Can you explain these vector fields?




Originally posted by Ambitwistor
... if you try to build coordinates out of those curves by constructing their tangent vectors
Doesn't the fact that you're trying to build coordinates by using vectors automatically tell you that it doesn't make sense? I thought that was the lesson I was learning on the sphere.




Originally posted by Ambitwistor
The frame field isn't a coordinate vector field; the vectors don't commute with each other.
I don't understand commuting/noncommuting vectors. The only experience I have with nontrivial commutation is in QM, but I only understand the concept with operators, not vectors. Can you explain?
lethe
#119
Dec13-03, 04:29 AM
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Originally posted by turin
From QM, I remember that the inner product was a bra multiplying a ket [...]

From classical mechanics, I remember that any vector has two different kinds of components and two different kinds of bases, contravariant and covariant. The contravariant basis is like arrows, and the covariant basis is like a set of planes (stack of pancakes). Taking a dot product (which I'm still trying to decide if this is really the same as an "inner product")
yes, dot product and inner product are synonymous.

of two vectors is summing the products of the contravariant components of one of them with the covariant components of the other. If the dot product is zero, I interpretted it as the arrow being parallel to the planes in the set of planes. The dot product is maximum when the arrow is perpendicular to the set of planes (so that it passes through the maximum number of planes).

Are you saying that my interpretations from classical mechanics are inappropriate? Should I think about it more in terms of the way I think about the inner product in QM?
in both cases, you are making heavy use of a certain fact without realizing it. when a vector space has an inner product, then there is an isomorphism between the vector space and the dual space. in the case of quantum mechanics, where your vector space is actually an infinite dimensional L2 Hilbert space, this is actually heavy theorem (Riesz), whereas in the finite dimensional case its rather trivial, but either way, its true.

so one way to think about an inner product is this: take two vectors in the vector space, feed them to the inner product machine, and get a number out. this is the better way to think of it, in my opinion.

another way to think of it, is, take two vectors, then use the isomorphism determined by the inner product to find the dual of one of the vectors, and then feed the other vector to that dual, getting a number out.

this latter is what you do in both descriptions you gave above. so in some sense, if you have an inner product, you can just not bother to distinguish between vectors and their dual. this is how you are thinking about vectors in classical: the only difference between a vector and a covector is whether it has raised or lowered indices. in QM, the only difference between a vector and a covector is whether it is a bra or a ket.

so, im not going to tell you that this way of thinking about vector spaces is wrong, but i have learned that it is very useful to, at the very least, remain aware of when you are using your inner product, and for what purposes. you never know when you may need to change your inner product, and then this isomorphism will change, and all the dual vectors you had previously written down will be something different. better still, don t use the isomorphism at all, and only write down a vector if you actually mean a vector, and a covector if you mean a covector.
lethe
#120
Dec13-03, 05:00 AM
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Originally posted by turin
But now it does make sense to me that the points in the manifold exist, period. The coordinate system is a way to keep track of them, like assigning an index to the elements of a set?
yeah, this sounds good. the points in the manifold are the points, and there may be many different coordinate systems that you can use to label the point, but we think of the point existing in some sense independently of which coordinate system we use.

Since the points are not discrete elements, we can't use an index (at least not like i = 1,2,3,...), so we use a continuous variable and call it a coordinate. Just like I could index the elements of a set differently, I can use a different coordinate.
yep

But now I am confused about why one needs two coordinates instead of one.
you need n coordinates, where n is the dimension of the manifold.

Is there no way to use just one coordinate to indicate points.
only if the manifold is 1 dimensional (a curve)
I am having trouble shifting my mind from elements of a set needing only one index and points in a manifold needing more than one coordinate.
if you label the elements of the set A with an index i, then you would use two indices (i,j) to label the elements of the set AxA. in the same sense, you need one real number to specify a point on the line (=R), 2 real numbers to specify a point on the plane (=RxR), etc.

i guess in a set theoretic sense, you don t actually need two indices. the cartesian product of two sets has the same cardinality, so you could count the elements of the new set with the same index, if you had a mind to, but this would muck up the other nice properties of the set, so it is much preferred to just stick to the double index concept.

(I am at this point going to assume that my analogy was correct that points:manifold::elements:set)
yes, its a good analogy
In a vector space, I've got the idea of linear independence dictating how many independent vectors I will need in my basis to span the vector space, so the dimension is clear. I don't see the same indicator in a manifold.
perhaps it would be useful to know the definition of a manifold. leaving out some technical details, the definition is this: an n-dimensional manifold is a space X such that for any point x in X, there is a neigborhood of x U and a continous invertible map with continuous inverse (homeomorphism) between U and Rn

in other words, if you look at a manifold up closely enough, it looks like Rn. these local homeomorphisms are the charts of the manifold, they are the coordinates.

Can you explain the collection of functions a bit?
perhaps it would be useful to see an example. a sphere can be written as the map [itex](u,v)\mapsto(\sin u\cos v,\sin u\sin v,cos v)[/itex]. here u and v are the coordinates. there are two of them because a sphere is a 2 dimensional manifold


But I thought that the tangent planes for any two points, no matter how close they are, have absolutely nothing to do with each other.
thats right. nothing at all (unless you like chroots approximations above. i don t.)

Unless I'm allowed to, in some sense, take the tangent plane with me, then I have to go from one to another. But, if adjacent tangent planes have nothing to do with each other, then this process seems disjoint, abrupt, or something. There just seems to be a discontinuity here, but I can't put my finger on it. I'll have to let this idea soak in my mind, too.
i m not sure what exactly you re saying here. i can assure you that, for example, if you have a smooth curve on the manifold, then this curve will have a tangent vector which is in the tangent space to the manifold at each point the curve passes through. furthermore, this tangent vector to the curve will vary from tangent space to tangent space, as the curve goes along, in a smooth way.

there is no discontinuity. perhaps it would be useful to know the following fact: a smooth manifold, together with the tangent space at each point, considered as one larger set, is also a smooth manifold. this manifold is called the tangent bundle.
lethe
#121
Dec13-03, 05:20 AM
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Originally posted by turin
What vector fields? I only know of vector fields like the electric field and such. Can you explain these vector fields?
a vector field is any mapping that assigns a vector to each point of the space. electric field is an example of a vector field on minkowski space, sure. you know how every vector space has a basis? a nice basis for cartesian space is i, j, and k. these are the unit vectors pointing along each coordinate. since there is a tangent space to every point in space, these form a basis at every point in space. hence they are vector fields (not just vectors). the vectors Ambitwistor are the same thing for polar coordinates.

this should be apparent to you. if you know that electric field is a vector field, and you know that electric field can be written [itex]\mathbf{E}=E_x\hat{\imath}+E_y\hat{\jmath}+E_z\hat{k}[/itex], so those i, j, and k are also vector fields. you can choose to work in polar coordinates, and thats all we are doing here.

Doesn't the fact that you're trying to build coordinates by using vectors automatically tell you that it doesn't make sense? I thought that was the lesson I was learning on the sphere.
points in the sphere are not vectors. never the less, given points in the sphere, you can calculate the tangent vectors, and given tangent vectors, you can calculate coordinates for points in the sphere.

I don't understand commuting/noncommuting vectors. The only experience I have with nontrivial commutation is in QM, but I only understand the concept with operators, not vectors. Can you explain?
vectors are differential operators. if you would like to understand why, well, i explained this concept at length once, when i started writing a tutorial for differential forms, here.

in short, there are various things you can do to define tangent vectors in an intrinsic way. a nice starting way is to define a tangent vector as an equivalence class of curves all of which have the same derivative at a point. with a little work, you see that this is canonically isomorphic to the space of first order differential operators on functions on the manifold. therefore i can simply say they are the same thing, and call a tangent vector on a manifold a differential operator.

this is tremendously useful, though unfamiliar.


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